ECS 289 / MAE 298, Network Theory Spring 2014 Problem Set # 1, Solutions. Problem 1: Power Law Degree Distributions (Continuum approach)

Transcription

1 ECS 289 / MAE 298, Network Theory Spring 204 Problem Set #, Solutions Problem : Power Law Degree Distributions (Continuum approach) Consider the power law distribution p(k) = Ak γ, with support (i.e., defined from) k = to k. a) Show that we must have γ > for this to be a properly defined probability distribution function (pdf). Recall a pdf must have two properties: ) p(k) 0 for all k, and 2) it must be normalized. Ak γ dk = A (γ ) k( γ+) ) If γ <, the limit (lim k k γ+ ). 2) If γ =, the prefactor A (γ ). = A [ ] ( lim k γ+ ) (γ ) k 3) If γ >, then the integral is finite: Ak γ dk = A. In other words, the integral (γ ) only converges if γ >. b) Solve for the normalization constant A. From (3) above, we find Ak γ = A (γ ) implies: A = (γ ). for γ >. Setting the integral equal to, c) Show that if < γ 2, the average value k diverges. k = (γ )k γ+ (γ ) dk = (γ 2) k( γ+2) ) If γ < 2, the limit (lim k k γ+2 ). 2) If γ = 2, the prefactor (γ ) (γ 2). (γ ) [ ] = ( lim k γ+2 ) (γ 2) k 3) If γ > 2, then the integral is finite: (γ )k γ+ dk = (γ ). In other words, the (γ 2) integral only converges if γ > 2.

2 d) Show that if 2 < γ 3, the average is finite, but the standard deviation diverges. k 2 = (γ )k γ+2 (γ ) dk = (γ 3) k( γ+3) ) If γ < 3, the limit (lim k k γ+3 ). 2) If γ = 3, the prefactor (γ ) (γ 3). (γ ) [ ] = ( lim k γ+3 ) (γ 3) k 3) If γ > 3, then the integral is finite: (γ )k γ+2 dk = (γ ). In other words, the (γ 3) integral only converges if γ > 3. Thus σ 2 = k 2 k 2 for γ 3. In general, for a power law of the form p(k) k γ, the moments k n diverge for all n (γ ). e) Plot p(k) = Ak γ, for k = to k = 00, 000 for γ = 3 (and properly normalize). Use matlab, R, or pen and paper, etc (but make sure to label axes clearly with values). The simple approach suggests that p(k) = 2k 3 since γ = 3. But the solution p(k) = 2k 3 is clearly problematic since it gives probability p(k = ) >. To properly normalize consider that this requires p(k) =, not p(k)dk = as used in part (b). We can solve for the normalization constant numerically using a simple program such as this C program which implements S = k 3 : 2

3 #i n c l u d e <s t d i o. h> #i n c l u d e <math. h> main ( ){ double x, S =0.; long i ; f o r ( i =; i $\ l e $ 00000; ++i ) S+=(pow( i, 3. ) ) ; p r i n t f ( %0.8 l f, S ) } The inverse normalization constant, A = S = (Note the discrete treatment in detail next gives this value). The correctly normalized plot is: f(x) e-06 e-08 e-0 e-2 e-4 e f) In a finite network with N nodes, what is the largest possible value of degree, k max, that can ever be observed? So can we ever have k in a finite network? In a finite network of N nodes, k max = (N ) (or N if we allow self-loops). Regardless, we can never observe k in a finite network. 3

4 Problem : Power Law Degree Distributions (Discrete approach) In most situations in networks, the k s take on only discrete values. We can do a similar treatment, where now we consider k discrete: p(k) = A k γ = A [ + 2 γ + 3 γ + 4 γ + ]. The sum in the brackets is known as the Riemann Zeta Function, RZ(γ). The value of RZ(γ), for many values of γ can be found in standard references (e.g., Mathworld, Wikipedia, etc). Commonly tabulated values are: RZ(0) = /2 RZ(/2).460 RZ() = RZ(3/2) 2.62 RZ(2) = π 2 /6.645 RZ(3).202 RZ(4) = π 4 / Clearly, for γ < 0, RZ(γ). Though the proof is beyond the scope of this assignment, the relevant information is that RZ(γ) < 0 for 0 γ <, RZ(), and RZ(γ) converges to a finite positive number for all γ >. (a) p(k) = A k γ = A RZ(γ). This is negative for 0 γ <, and it diverges for γ =. Thus for p(k) to converge to a finite positive number requires γ >. (b) Normalization constant, A = /RZ(γ). (c) k = kp(k) = [ A ] 2 γ 3 γ 4 γ = A [ ] = A RZ(γ ) = RZ(γ )/RZ(γ), 2 γ 3 γ 4 γ which is negative or diverges for γ or equivalently, γ 2. Thus k only converges to a positive finite number for γ > 2. (d) Likewise, k 2 = A RZ(γ 2) = RZ(γ 2)/RZ(γ), which is negative for 2 γ < 3 and diverges for γ = 3. Thus σ 2 = k 2 k 2 = RZ(γ 2)/RZ(γ) [RZ(γ )/RZ(γ)] 2 only converges to a positive finite number for γ > 3. (e) For plot, see page 3. (f) In a finite network of N nodes, k max = (N ) (or N if we allow self-loops). Regardless, we can never observe k in a finite network. 4

5 Problem 2: Random walk on a network (a) Let A ij = in the following matrix denote the presence of a link from node j to node i and A ij = 0 denote the absence of a link. The adjacency matrix is given by A = 0 0 () (b) In order to obtain the steady state distribution we normalize the adjacency matrix given above to obtain the transition matrix T. Entry T ij denotes the probability of going from node j to node i in the next time step T = (2) The steady state probability corresponds to the eigenvector of T with eigenvalue. Using a standard software package we find that the eigenvector that corresponds to the eigenvalue of is π = (3)

6 Note that the above eigenvector is normalized such that the sum of the square of the elements is equal to. Renormalizing such that the sum of the terms is we obtain π = 0.4 (4) (c) For any undirected graph, the steady state probability of finding a random walker is directly proportional to the degree of the node. Therefore, 3 π = (5) 2 2 6